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Free_Kalibri [48]
3 years ago
13

120 grams of calcium nitrite Ca(NO2)2 is dissolved in a 240 mL solution. What is the molarity of the solution???

Chemistry
1 answer:
expeople1 [14]3 years ago
5 0

Hey there!:

Molar mass Ca(NO2)2 = 132.089 g/mol

Mass of solute = 120 g

Number of moles:

n = mass of solute / molar mass

n = 120 / 132.089

n = 0.0009084 moles of Ca(NO2)2

Volume in liters of solution :

240 mL / 1000 => 0.24 L

Therefore:

Molarity = number of moles / volume of solution

Molarity = 0.0009084 / 0.24

Molarity = 0.003785 M

Hope that helps!

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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
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Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
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Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

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The short of it is you would create salt.

Explanation:

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Given all option is correct but most appropriate option is a.

So the option a is correct

(a) The coefficient of performance of an irreversible refrigeration cycle is always less than the coefficient of performance of a reversible refrigeration cycle when both exchange energy by heat transfer with the same two reservoirs.

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